Codeforces Round #411 (Div. 2)

A

#include <cstdio>#include <cstring>#include <cmath>#include <sstream>#include <iostream>#include <algorithm>#include <vector>#include <stack>#include <queue>#include <string>#include <map>#include <set>#include <utility>using namespace std;#define LL long long#define pb push_back#define pill pair<int, int>#define mk make_pair#define mst(a, b) memset(a, b, sizeof a)#define REP(i, x, n)for(int i = x; i < n; ++i)const int qq = 1e5 + 10;int n, m, k;int num[qq];int main(){int l, r;scanf("%d%d", &l, &r);if(l == r){printf("%d\n", l);return 0;}printf("2\n");return 0;}

B

#include <cstdio>#include <cstring>#include <cmath>#include <sstream>#include <iostream>#include <algorithm>#include <vector>#include <stack>#include <queue>#include <string>#include <map>#include <set>#include <utility>using namespace std;#define LL long long#define pb push_back#define pill pair<int, int>#define mk make_pair#define mst(a, b) memset(a, b, sizeof a)#define REP(i, x, n)for(int i = x; i < n; ++i)const int qq = 1e5 + 10;int n, m, k;int num[qq];int main(){scanf("%d", &n);int t = 0;for(int i = 1; i <= n; ++i){if(t == 0)putchar('a');if(t == 1)putchar('a');if(t == 2)putchar('b');if(t == 3)putchar('b');t++;if(t == 4)t = 0;}puts("");return 0;}

C

#include <cstdio>#include <cstring>#include <cmath>#include <sstream>#include <iostream>#include <algorithm>#include <vector>#include <stack>#include <queue>#include <string>#include <map>#include <set>#include <utility>using namespace std;#define LL long long#define pb push_back#define pill pair<int, int>#define mk make_pair#define mst(a, b) memset(a, b, sizeof a)#define REP(i, x, n)for(int i = x; i < n; ++i)const int qq = 1e5 + 10;int n, m, k;int num[qq];int main(){scanf("%d", &n);if(n == 1 || n == 2){printf("0\n");return 0;}printf("%d\n", (n - 1) / 2);return 0;}

D

题意：ab能变成bba，询问最多可以操作多少次。

思路：从结果来看我们知道最后一定是bbbb....aaa....，可知其实操作就是将所有的a向右边移动，要使得操作次数最多我们可以从最右端的a开始向右端挪动，每次将一个a移动到不能再移动，此时b的数量变成原来的两倍。然后模拟即可

#include <cstdio>#include <cstring>#include <cmath>#include <sstream>#include <iostream>#include <algorithm>#include <vector>#include <stack>#include <queue>#include <string>#include <map>#include <set>#include <utility>using namespace std;#define LL long long#define pb push_back#define pill pair<int, int>#define mk make_pair#define mst(a, b) memset(a, b, sizeof a)#define REP(i, x, n)for(int i = x; i < n; ++i)const int qq = 1e6 + 10;const int mod = 1e9 + 7;int n, m, k;char st[qq];int main(){scanf("%s", st);LL ans = 0;LL b = 0;int len = (int)strlen(st) - 1;for(int i = len; i >= 0; --i){if(b == 0 && st[i] == 'a')continue;if(st[i] == 'b')b = (b + 1) % mod;else if(st[i] == 'a')ans = (ans + b + mod) % mod, b = (2ll * b + mod) % mod;}printf("%lld\n",(ans + mod) % mod);return 0;}

E

题意：有n个节点，每一个节点上有si个ice cream，现在要让ice cream组成一个图G， ice cream中任意两个点存在边的条件是属于同一个节点(n中的节点)，换句话说也就是每一个节点上的点都组成一个完全图，现在问你最小染色数以及方案

思路：我一直以为后面给的树是完全没有用的，恩。。。      其实是漏了这么一句话

Vertices which have the i-th (1 ≤ i ≤ m) type of ice cream form a connected subgraph

就是拥有 i-th (1 ≤ i ≤ m) type of ice cream的节点组成一个连通图，k个节点，k - 1条边的连通图。

如果我们对每一个节点去进行染色的话， 从1开始递进染色，这里会有很多意想不到的情况，具体可以自己试试然后看看自己错的那组数据，所以题目这么说就是为了把我们的思路转向所给的树上，对树来进行dfs染色。

#include <bits/stdc++.h>using namespace std;#define LL long long#define pb push_back#define mst(a, b)memset(a, b, sizeof a)#define REP(i, x, n)for(int i = x; i < n; ++)const int qq = 5e5 + 10;vector<int> T[qq], G[qq];map<int, bool> mp;int color[qq];int maxn = 1;void Dfs(int u, int fa){mp.clear();for(int v : G[u]){if(color[v])mp[color[v]] = true;}int cnt = 1;for(int v : G[u]){if(!color[v]){while(mp[cnt] == true)cnt++;mp[cnt] = true;color[v] = cnt;}}maxn = max(maxn, cnt);for(int v : T[u]){if(v == fa)continue;Dfs(v, u);}}int main(){int n, m;scanf("%d%d", &n, &m);for(int i = 1; i <= n; ++i){int k;scanf("%d", &k);for(int j = 0; j < k; ++j){int x;scanf("%d", &x);G[i].pb(x);}}int a, b;for(int i = 1; i < n; ++i){scanf("%d%d", &a, &b);T[a].pb(b), T[b].pb(a);}Dfs(1, 0);for(int i = 1; i <= m; ++i)if(!color[i])color[i] = 1;printf("%d\n", maxn);for(int i = 1; i <= m; ++i){printf("%d ", color[i]);}puts("");return 0;}